a) \(\left(\frac{1}{16}\right)^{200}\) và \(\left(\frac{1}{2}\right)^{1000}.\)

Ta có:

\(\left(\frac{1}{16}\right)^{200}=\left[\left(\frac{1}{2}\right)^4\right]^{200}=\left(\frac{1}{2}\right)^{800}.\)

\(\left(\frac{1}{2}\right)^{1000}.\)

Vì \(800< 1000\) nên \(\left(\frac{1}{2}\right)^{800}< \left(\frac{1}{2}\right)^{1000}.\)

\(\Rightarrow\left(\frac{1}{16}\right)^{200}< \left(\frac{1}{2}\right)^{1000}.\)

Chúc bạn học tốt!

câu d thì dễ

Thế giúp vs bn ơi

mk bn lớp mấy

a) Ta có \(0,625^{200}=\left(\dfrac{5}{8}\right)^{200}\) và \(0,5^{1000}=\left(\dfrac{1}{2}\right)^{1000}=\left(\dfrac{1}{2}\right)^{5.200}\) \(=\left[\left(\dfrac{1}{2}\right)^5\right]^{200}\) \(=\left(\dfrac{1}{32}\right)^{200}\). Mà hiển nhiên \(\left(\dfrac{5}{8}\right)^{200}>\left(\dfrac{1}{32}\right)^{200}\) nên suy ra \(0,625^{200}>0,5^{1000}\)

b) Ta thấy \(\left(-32\right)^{27}< 0\) trong khi \(\left(-27\right)^{32}>0\) nên đương nhiên \(\left(-32\right)^{27}< \left(-27\right)^{32}\)

c) Ta thấy \(-\dfrac{3}{2}>-2\) nên \(\left(-\dfrac{3}{2}\right)^5>\left(-2\right)^5\)

#)Giải :

Ta có : 18³⁹ > 16³⁹ = 2¹⁵⁶ > 2¹³⁵ = 32²⁷

            18³⁹ > 32²⁷ => (-18)³⁹ < (-32)²⁷

Vậy : (-18)³⁹ < (-32)²⁷

         #~Will~be~Pens~#

Trả lời : 

Ta có: 32²⁷=(2⁵)²⁷=2¹³⁵<2¹⁵⁶=2^4.39=16³⁹<18³⁹

=>(-32)²⁷>(-18)³⁹

Vậy (-32)²⁷>(-18)³⁹

\(\downarrow\)

Ta có \(32^{27}=\left(2^5\right)^{27}=2^{135}< 2^{136}=\left(2^4\right)^{39}=16^{39}< 18^{39}\)

\(\Rightarrow\left(-32\right)^{27}>\left(-18\right)^{39}\)

a, Có : (1/60)^200 = [(1/2)^4]^200 = (1/2)^800

Vì 0 < 1/2 < 1 nên (1/2)^800 > (1/2)^1000

=> (1/16)^200 > (1/2)^1000

Tk mk nha

a) \(\left(\frac{1}{16}\right)^{200}=\left(\frac{1}{2}\right)^{800}< \left(\frac{1}{2}\right)^{1000}\)

a) \(\left(\frac{1}{16}\right)^{200}=\frac{1}{16^{200}}\)

\(\left(\frac{1}{2}\right)^{1000}=\frac{1}{2^{1000}}\)

có : \(16^{200}=\left(2^4\right)^{200}=2^{800}\)

ta thấy \(2^{800}< 2^{1000}\)

\(\Rightarrow16^{200}< 2^{1000}\)

\(\Rightarrow\frac{1}{16^{200}}>\frac{1}{2^{1000}}\)

\(\Rightarrow\left(\frac{1}{16}\right)^{200}>\left(\frac{1}{2}\right)^{100}\)

Câu 1:

\(A=27^2.32^3=\left(3^3\right)^2.\left(2^5\right)^3=3^6.2^{15}\)

\(B=6^{16}=2^{16}.3^{16}\)

Từ \(\hept{\begin{cases}2^{15}< 2^{16}\\3^6< 3^{16}\end{cases}\Leftrightarrow2^{15}.3^6< 2^{16}.3^{16}\Leftrightarrow}A< B\)

Câu 2:

\(A=1+2+2^2+2^3+...+2^{2016}\)

<=>\(2A=2\left(1+2+2^2+2^3+...+2^{2016}\right)\)

<=>\(2A=2+2^2+2^3+2^4...+2^{2017}\)

<=>\(2A-A=\left(2+2^2+2^3+2^4+...+2^{2017}\right)-\left(1+2+2^2+2^3+...+2^{2016}\right)\)

<=>\(A=2^{2017}-1< 2^{2017}=B\)

Vậy A<B

muốn viết dấu mũ như thế kia thì viết thế nào hả bạn ?

A = 27² . 32³                            B = 6¹⁶

A = ( 3³ )² . ( 2⁵ )³                        B = ( 2 .3 )¹⁶

A = 3⁶ . 2¹⁵                               B = 2¹⁶ . 3¹⁶

      ta thấy 3⁶ < 3¹⁶ ; 2¹⁵ < 2¹⁶

\(\Rightarrow A< B\)

chúc bạn học giỏi ^^

\(\left(\dfrac{1}{16}\right)^{200}< \left(\dfrac{1}{2}\right)^{1000}\)